# How do you solve the system -x-5y-5z=2, 4x-5y+4z=19, and x+5y-z=-20?

Jan 23, 2017

Verify that the determinant of the coefficient matrix is not zero.

If so, then create and augmented matrix by appending of column of constants.

Perform elementary row operations, until solved.

#### Explanation:

Let A equal the coefficient matrix :

A = [ (-1,-5,-5), (4,-5,4), (1,5,-1) ]

Verify that the determinant is not zero:

|A| = |(-1,-5,-5), (4,-5,4), (1,5,-1) | = -140

NOTE: I wrote all of the steps to compute the determinant but it made the write up too long. Please read the reference link. The determinant is not zero, therefore there is a unique solution.

Let B = the column of constants:

$B = \left[\begin{matrix}2 \\ 19 \\ - 20\end{matrix}\right]$

Write augmented matrix by appending B to A:

A|B = [ (-1,-5,-5,|,2), (4,-5,4,|,19), (1,5,-1,|,-20) ]

Perform elementary row operations until the left is an identity matrix :

${R}_{1} + {R}_{3} \to {R}_{3}$

[ (-1,-5,-5,|,2), (4,-5,4,|,19), (0,0,-6,|,-18) ]

${R}_{3} / - 6$

[ (-1,-5,-5,|,2), (4,-5,4,|,19), (0,0,1,|,3) ]

$- 4 {R}_{3} + {R}_{2} \to {R}_{2}$

[ (-1,-5,-5,|,2), (4,-5,0,|,7), (0,0,1,|,3) ]

$5 {R}_{3} + {R}_{1} \to {R}_{1}$

[ (-1,-5,0,|,17), (4,-5,0,|,7), (0,0,1,|,3) ]

$4 {R}_{1} + {R}_{2} \to {R}_{2}$

[ (-1,-5,0,|,17), (0,-25,0,|,75), (0,0,1,|,3) ]

${R}_{2} / - 25$

[ (-1,-5,0,|,17), (0,1,0,|,-3), (0,0,1,|,3) ]

$5 {R}_{2} + {R}_{1} \to {R}_{1}$

[ (-1,0,0,|,2), (0,1,0,|,-3), (0,0,1,|,3) ]

$\left(- 1\right) {R}_{1}$

[ (1,0,0,|,-2), (0,1,0,|,-3), (0,0,1,|,3) ]

We have an identity matrix on the left, therefore, the right contains the values of the variables:

$x = - 2 , y = - 3 , z = 3$

check:

$- x - 5 y - 5 z = 2$
$4 x - 5 y + 4 z = 19$
$x + 5 y - z = - 20$

$- \left(- 2\right) - 5 \left(- 3\right) - 5 \left(3\right) = 2$
$4 \left(- 2\right) - 5 \left(- 3\right) + 4 \left(3\right) = 19$
$\left(- 2\right) + 5 \left(- 3\right) - \left(3\right) = - 20$

$2 = 2$
$19 = 19$
$- 20 = - 20$

This checks.